Proving a statement about real numbers












2












$begingroup$



If $x, y in Bbb R$ and $x<y+epsilon$ for every $epsilon >0$, then $x<y$.




Okay so I went about this by proving the contrapositive.



Proof: Let $x,yinBbb R$ and let $epsilon>0$. Suppose $xge y$. Take $epsilon = x-y$. This implies that $x=y+epsilon$, as needed.



Is this a valid proof or not?






real-analysis proof-verification real-numbers







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The conclusion should be $xleq y$.
    $endgroup$
    – gamma
    3 hours ago










  • $begingroup$
    If $$epsilon=x-y$$ then your epsilon would be negative!
    $endgroup$
    – Dr. Sonnhard Graubner
    3 hours ago










  • $begingroup$
    Why would it be negative if $x ge y$?
    $endgroup$
    – Ash
    3 hours ago






  • 1




    $begingroup$
    1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
    $endgroup$
    – fleablood
    3 hours ago












  • $begingroup$
    I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
    $endgroup$
    – fleablood
    3 hours ago
















2












$begingroup$



If $x, y in Bbb R$ and $x<y+epsilon$ for every $epsilon >0$, then $x<y$.




Okay so I went about this by proving the contrapositive.



Proof: Let $x,yinBbb R$ and let $epsilon>0$. Suppose $xge y$. Take $epsilon = x-y$. This implies that $x=y+epsilon$, as needed.



Is this a valid proof or not?






real-analysis proof-verification real-numbers







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The conclusion should be $xleq y$.
    $endgroup$
    – gamma
    3 hours ago










  • $begingroup$
    If $$epsilon=x-y$$ then your epsilon would be negative!
    $endgroup$
    – Dr. Sonnhard Graubner
    3 hours ago










  • $begingroup$
    Why would it be negative if $x ge y$?
    $endgroup$
    – Ash
    3 hours ago






  • 1




    $begingroup$
    1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
    $endgroup$
    – fleablood
    3 hours ago












  • $begingroup$
    I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
    $endgroup$
    – fleablood
    3 hours ago














2












2








2





$begingroup$



If $x, y in Bbb R$ and $x<y+epsilon$ for every $epsilon >0$, then $x<y$.




Okay so I went about this by proving the contrapositive.



Proof: Let $x,yinBbb R$ and let $epsilon>0$. Suppose $xge y$. Take $epsilon = x-y$. This implies that $x=y+epsilon$, as needed.



Is this a valid proof or not?






real-analysis proof-verification real-numbers







share|cite|improve this question











$endgroup$





If $x, y in Bbb R$ and $x<y+epsilon$ for every $epsilon >0$, then $x<y$.




Okay so I went about this by proving the contrapositive.



Proof: Let $x,yinBbb R$ and let $epsilon>0$. Suppose $xge y$. Take $epsilon = x-y$. This implies that $x=y+epsilon$, as needed.



Is this a valid proof or not?






real-analysis proof-verification real-numbers




real-analysis proof-verification real-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









J.G.

29.4k22846




29.4k22846










asked 3 hours ago









AshAsh

234




234








  • 3




    $begingroup$
    The conclusion should be $xleq y$.
    $endgroup$
    – gamma
    3 hours ago










  • $begingroup$
    If $$epsilon=x-y$$ then your epsilon would be negative!
    $endgroup$
    – Dr. Sonnhard Graubner
    3 hours ago










  • $begingroup$
    Why would it be negative if $x ge y$?
    $endgroup$
    – Ash
    3 hours ago






  • 1




    $begingroup$
    1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
    $endgroup$
    – fleablood
    3 hours ago












  • $begingroup$
    I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
    $endgroup$
    – fleablood
    3 hours ago














  • 3




    $begingroup$
    The conclusion should be $xleq y$.
    $endgroup$
    – gamma
    3 hours ago










  • $begingroup$
    If $$epsilon=x-y$$ then your epsilon would be negative!
    $endgroup$
    – Dr. Sonnhard Graubner
    3 hours ago










  • $begingroup$
    Why would it be negative if $x ge y$?
    $endgroup$
    – Ash
    3 hours ago






  • 1




    $begingroup$
    1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
    $endgroup$
    – fleablood
    3 hours ago












  • $begingroup$
    I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
    $endgroup$
    – fleablood
    3 hours ago








3




3




$begingroup$
The conclusion should be $xleq y$.
$endgroup$
– gamma
3 hours ago




$begingroup$
The conclusion should be $xleq y$.
$endgroup$
– gamma
3 hours ago












$begingroup$
If $$epsilon=x-y$$ then your epsilon would be negative!
$endgroup$
– Dr. Sonnhard Graubner
3 hours ago




$begingroup$
If $$epsilon=x-y$$ then your epsilon would be negative!
$endgroup$
– Dr. Sonnhard Graubner
3 hours ago












$begingroup$
Why would it be negative if $x ge y$?
$endgroup$
– Ash
3 hours ago




$begingroup$
Why would it be negative if $x ge y$?
$endgroup$
– Ash
3 hours ago




1




1




$begingroup$
1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
$endgroup$
– fleablood
3 hours ago






$begingroup$
1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
$endgroup$
– fleablood
3 hours ago














$begingroup$
I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
$endgroup$
– fleablood
3 hours ago




$begingroup$
I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
$endgroup$
– fleablood
3 hours ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

The statement you want to prove is not true.



Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You're absolutely correct, thank you.
    $endgroup$
    – Ash
    3 hours ago



















2












$begingroup$

The proof is invalid for the simple fact that the statement is false, so you cannot prove it.



The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.



Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    1) The statement is false.



    Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.



    2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.



    Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.



    3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.



    4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      The statement you want to prove is not true.



      Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        You're absolutely correct, thank you.
        $endgroup$
        – Ash
        3 hours ago
















      3












      $begingroup$

      The statement you want to prove is not true.



      Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        You're absolutely correct, thank you.
        $endgroup$
        – Ash
        3 hours ago














      3












      3








      3





      $begingroup$

      The statement you want to prove is not true.



      Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.






      share|cite|improve this answer









      $endgroup$



      The statement you want to prove is not true.



      Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 3 hours ago









      TheSilverDoeTheSilverDoe

      3,418112




      3,418112












      • $begingroup$
        You're absolutely correct, thank you.
        $endgroup$
        – Ash
        3 hours ago


















      • $begingroup$
        You're absolutely correct, thank you.
        $endgroup$
        – Ash
        3 hours ago
















      $begingroup$
      You're absolutely correct, thank you.
      $endgroup$
      – Ash
      3 hours ago




      $begingroup$
      You're absolutely correct, thank you.
      $endgroup$
      – Ash
      3 hours ago











      2












      $begingroup$

      The proof is invalid for the simple fact that the statement is false, so you cannot prove it.



      The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.



      Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        The proof is invalid for the simple fact that the statement is false, so you cannot prove it.



        The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.



        Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          The proof is invalid for the simple fact that the statement is false, so you cannot prove it.



          The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.



          Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.






          share|cite|improve this answer









          $endgroup$



          The proof is invalid for the simple fact that the statement is false, so you cannot prove it.



          The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.



          Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          egregegreg

          184k1486205




          184k1486205























              0












              $begingroup$

              1) The statement is false.



              Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.



              2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.



              Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.



              3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.



              4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                1) The statement is false.



                Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.



                2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.



                Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.



                3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.



                4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  1) The statement is false.



                  Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.



                  2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.



                  Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.



                  3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.



                  4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.






                  share|cite|improve this answer









                  $endgroup$



                  1) The statement is false.



                  Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.



                  2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.



                  Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.



                  3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.



                  4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  fleabloodfleablood

                  72.1k22687




                  72.1k22687






























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